Use Euler;s Theorem to find all incongruent solutions of each congruence below.

9x is congruent to 21 mod 25.

OK, I know this is an easy computational but my anwser still does not math the books. can someone spot the error:

1) i found the inverse of 9 modulo 25 to be 9^22. multiply both side and get

2) x is congruent to 9^22 times 21 mod 25

3) so then i get x is congruent to 7 mod 25.

and i get the anwser of 4.

I dont know where are my mistakes?

Jul 29th 2008, 12:05 AM

Moo

Hello,

Quote:

1) i found the inverse of 9 modulo 25 to be 9^22. multiply both side and get

false
9^23 is not congruent to 1 mod 25

Quote:

2) x is congruent to 9^22 times 21 mod 25

correct reasoning

Quote:

3) so then i get x is congruent to 7 mod 25.

9^22 times 21 mod 25 is congruent to 1 mod 25

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Anyway, this looks naaasty ! In order to find the multiplicative inverse of the number, use the extended Euclidian algorithm.

But first of all, let's play a little with the equation :
That is to say .
Since 3 & 25 are coprime, divides 25, that is to say .
In further exercises, you can directly "simplify" by 3, since it divides 9x, 21, but not 25. If it were 24, you would have had to divide by 3 too.

Now, find the inverse of 3 mod 25. Since 3 and 25 are coprime, there exists a such that . But since 3 and 25 are coprime, we can also say that their gcd is 1. And we know that at a moment, we'll get their gcd as a remainder in the Euclidian algorithm. So let's do it and you will see that thereafter.

thus , .
That is the inverse of 3.

So (Wink)

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Now, if you have to use Euler's totient function (phi) :

(in general, if p is prime, then ). Thus for any number coprime with 25, for example 9, we have

. Therefore is the inverse of 9 mod 25 !

Does this help ?

Jul 29th 2008, 05:07 AM

ThePerfectHacker

There is an easy way to get the inverse.
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